Integrals Test

Result

Integrals Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \int _{ 0 }^{ { \pi }^{ 2 } }{ \dfrac { \sin { \sqrt { x } } }{ \sqrt { x } } } dx$$ is equal to

A
$$2$$

B
$$1$$

C
$$1/2$$

D
$$4$$

Solution
Question 2
1 / -0

The integral $$\int _{ 2a/4 }^{ a/2 }{ (2\quad cosecx{ ) }^{ 17 } } $$ dx is equal to:

A
$$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }+{ e }^{ -u }{ ) }^{ 16 } } du$$

B
$$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }+{ e }^{ -u }{ ) }^{ 17 } } du$$

C
$$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }-{ e }^{ -u }{ ) }^{ 17 } } du$$

D
$$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }-{ e }^{ -u }{ ) }^{ 16 } } du$$
Question 3
1 / -0

Solve : $$\int^1_{0^+} \dfrac{x^x(x^{2x} +1)(lnx +1)}{x^{4x}+1} dx$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution
Question 4
1 / -0

$$\displaystyle \int \frac { e ^ { x } ( 1 + \sin x ) } { 1 + \cos x } d x =$$

A
$$e ^ { x } \tan x + c$$

B
$$e ^ { x } \sec ^ { 2 } \frac { x } { 2 } + c$$

C
$$e ^ { x } \tan \frac { x } { 2 } + c$$

D
$$\frac { 1 } { 2 } e ^ { x } \tan \frac { x } { 2 } + c$$

Solution

Let $$I=\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$$

$$=\displaystyle\int e^x\left(\dfrac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$$

$$=\displaystyle\int e^x\left(\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin\dfrac{x}{2}.\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$$

$$=\displaystyle\int e^x\left(\dfrac{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2}{2\cos^2\dfrac{x}{2}}\right)$$

$$=\displaystyle\int \dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$$

$$=\displaystyle\int\dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}+\dfrac{\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$$

$$=\displaystyle\int\dfrac{e^x}{2}\left(\tan\dfrac{x}{2}+1\right)^2$$

$$=\displaystyle\int\dfrac{e^x}{2}\left(\tan^2\dfrac{x}{2}+1+2\tan\dfrac{x}{2}\right)$$

$$=\displaystyle\int\dfrac{e^x}{2}\left(\sec^2\dfrac{x}{2}+2\tan\dfrac{x}{2}\right)$$

$$=\displaystyle\int e^x\left(\dfrac{1}{2}.\sec^2\dfrac{x}{2}+\dfrac{2}{2}\tan\dfrac{x}{2}\right)$$

$$=\displaystyle\int e^x\left(\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)$$

Above is in the form of $$e^x[f(x)+f'(x)]dx=e^xf(x)+c$$
$$=e^x\tan\dfrac{x}{2}+c$$

$$\therefore$$ $$\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$$ $$=e^x\tan\dfrac{x}{2}+c$$
Question 5
1 / -0

The value of the integral $$\int _{ 0 }^{ 1 }{ \sqrt { \frac { 1-x }{ 1+x } } } $$ dx is

A
$$\frac { \pi }{ 2 } +1$$

B
$$\frac { \pi }{ 2 } -1$$

C
-1

D
1

Solution
Question 6
1 / -0

Let $$A = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 }$$ dt, then the value of $$\int _ { 0 } ^ { 1 } \frac { t e ^ { t ^ { 2 } } } { t ^ { 2 } + 1 } d t$$ is:-

A
$$A ^ { 2 }$$

B
$$\frac { 1 } { 2 } A$$

C
2$$A$$

D
$$\frac { 1 } { 2 } A ^ { 2 }$$

Solution
Question 7
1 / -0

The value of the integral $$\int _{ -\pi /2 }^{ \pi /2 }{ \left[ { x }^{ 2 }+log\frac { \pi -x }{ \pi +x } \right] } $$ cos x dx is

A
0

B
$$\frac { { \pi }^{ 2 } }{ 2 } -4$$

C
$$\frac { { \pi }^{ 2 } }{ 2 } +4$$

D
$$\frac { { \pi }^{ 2 } }{ 2 } $$

Solution
Question 8
1 / -0

$$\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { x + 1 } + \sqrt { x } } d x =$$

A
$$\frac { 4 } { 3 } ( \sqrt { 2 } + 1 )$$

B
$$\frac { 4 } { 3 } ( \sqrt { 2 } - 1 )$$

C
$$\frac { 3 } { 4 } ( \sqrt { 2 } - 1 )$$

D
$$\frac { 3 } { 4 } ( \sqrt { 2 } - 2 )$$

Solution
Question 9
1 / -0

The value of the integral $$\int _ { - \pi } ^ { \pi } ( \cos p x - \sin q x ) ^ { 2 } d x$$ where $$p , q$$ are integers, is equal to:-

A
$$- \pi$$

B
0

C
$$\pi$$

D
2p

Solution
Question 10
1 / -0

$$\int _{ -1 }^{ 1/2 }{ \dfrac { { e }^{ x }\left( 2-{ x }^{ 2 } \right) dx }{ \left( 1-x \right) \sqrt { 1-{ x }^{ 2 } } } } $$ is equal to

A
$$\dfrac { \sqrt { e } }{ 2 } \left( \sqrt { 3 } +1 \right) $$

B
$$\dfrac { \sqrt { 3e } }{ 2 } $$

C
$$\sqrt { 3e } $$

D
$$\sqrt { \dfrac { e }{ 3 } } $$

Solution